Question: A certain circle can be represented by the following equation. $x^2+y^2+16x-14y+49=0$ What is the center of this circle ? $($
Explanation: The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2+16x-14y+49&=0\\\\ x^2+y^2+16x-14y&=-49\\\\ (x^2+16x)+(y^2-14y)&=-49 \text{(rearrange terms)}\\\\ (x^2+16x{+64})+(y^2-14y{+49})&=-49{+64}{+49}\end{aligned}$ Notice that we must add ${64}$ and ${49}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 64 and 49?] Writing the equation in standard form $\begin{aligned}(x^2+16x{+64})+(y^2-14y{+49})&=-49{+64}{+49}\\\\ (x+8)^2+(y-7)^2&=64\\\\ (x-(-8))^2+(y-7)^2&=8^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(-8,7)$ and has a radius of $8$ units. Summary The circle is centered at $(-8,7)$. The circle has a radius of $8$ units.